scheme: can’t find a readable default for option –band.
searched for file all.com in these directories:
/usr/lib/mit-scheme

Inconsistency detected.

I tried nearly ever combination of symbolic linking from the Applications directory into usr/lib and I even created the infamous usr/local/lib despite its original non-existence.

One source suggested I set the path attribute in my profile. Since I had been copying willy-nilly the symbolic links to just about every location in /bin, /usr/bin/ and usr/lib, nothing made sense. I eliminated those references to start clean and simply updated my .profile with the following:

export MITSCHEME_LIBRARY_PATH=/Applications/mit-scheme.app/Contents/Resources

That will allow the Scheme executable to find the library when it runs. But how can you have it get recognized in the terminal? You actually do need a copy of the Scheme binary in there, or at least a symbolic link. So, just cd into the /usr/bin directory and then make the symlink.

sudo ln -s /Applications/mit-scheme.app/Contents/Resources/mit-scheme scheme

If you try entering scheme into the terminal now, you’ll either get nothing (e.g. it’s not a proper bash command) or there’s inconsistency detected. What you’ll have to do is exit completely from the terminal. The paths have to update and restarting the terminal is the easy way to do it. After the restart though, you’ll see that scheme should work just fine now when called from the terminal on OSX.

public static void main(String[] args) {
	int[] arr = new int[] {2804, 2374, 6382, 1899};
	
	int index = 0, largest = arr[0];
	for (int i = 1; i < arr.length; i++) {
		(if largest > arr[i]) {
			index = i;
			largest = arr[i];
		}
	}
}

It’s your typical searching for the largest value loop. It’s nothing special, but then again, it doesn’t work.

At least the error message is logical. No, not really.

the left-hand side of an assignment must be a variable

By studying very carefully and fully realizing that I am still in scheme mode, I notice that there is a parenthesis before the if-statement instead after the if keyword itself. Had I been helping someone else with this code and they showed me that error message along with this, would I have known it was a misplaced parenthesis? No way. I’d eliminate one possibility at a time by removing each successive layer of expressions and assignments until it stopped. I might even go so far as to retype the entire thing. It’s subtle.

It also turns out that if you try to assign a value to a method, you’ll cause this error too.

Happy language switching!

Here’s an example. Look at this code. I dare you even to try running it. You’ll end up with an error.

public static void main(String[] args) {
	int magic = 10;
	if (magic < 20);
	{
		System.out.println("weak magic");
	} 
	else
	{
		System.out.println("strong magic");
	}
}	

The error is of course, uncommunicative of the true problem. Did you spot the error yet?

syntax error on token “else”, delete this token

The problem here isn’t the else or even the if. No, the problem is a semi-colon. Seriously. Take a look at the if-statement expression. The oddity comes about when you actually see the semi-colon attached to the end of it. And because the if-statement ended there, the else has no preceding pair so it has no choice but to error out.

This example is actually a clear cut case of putting a semi-colon after an if-statement expression. Imagine you didn’t have an error to remind you what you just did. You might just have a plain statement and proper following brackets.

public static void main(String[] args) {
	int magic = 10;
	if (magic < 20); {
		System.out.println("weak magic");
	}  
}

Without the else to throw an error message up, how easy would this be to spot? It’s tricky. What if your if-case was a rare edge case condition that you were hard pressed to demonstrate in testing but you theoretically knew it existed? You’d almost never know this wasn’t work until your nuclear reactor began to melt down.

The moral of story: kids will do what you tell them and mind you semi-colons at the end of constructs.

Duplicate local variable

Unfortunately, I didn’t grab his erring source code earlier, so I can’t give the true real world source. Nevertheless, my own crafted code should illuminate the problem sufficiently.

public static void main(String[] args) {
	
	int magic = 10;
	if ( 10 < 11 ) {
		int magic = 12;
		magic++;
	}
	
}

Imagine you had a variable declaration and initialization at the top of some method. Then later you attempt to create another variable with the same name as before, but inside of a construct like a for loop or if. Unwittingly, you have attempted to redeclare an already declared variable.

Why doesn’t the IDE or compiler know that it’s a syntax error in that case, and that you obviously cannot redeclare variables? Java’s just funny like that, but I suspect it has something to do with the scoping rules in Java. The magic variable inside of the if-statement should be separate, and it is telling you it’s a duplicate, not that it isn’t allowed. So I suppose that’s helpful.

In short, the easy fix is to rename the inner-conflicting variable. So the magic that’s in the if-statement could be magic2 and anything that references it in the if should also be updated. That’s clear enough. The other possibility is to remove the int declaration. Imagine you were simply setting value instead of declaring. It’s an easy mistake to make, especially when you’ve a bunch of declarations repeated.

Happy deduplicating!

testing.c:(.text+0×156): undefined reference to `pow’
testing.c:(.text+0×174): undefined reference to `pow’
testing.c:(.text+0x18e): undefined reference to `pow’

#include <stdio.h>
#include <stdlib.h>
#include <math.h> 
#include <time.h>

My code was modestly simple. The variables were previously defined and perfectly in order.

      double left = a * pow(b, 3);
      double right = pow(c, 2) - pow(d, e);

Apparently, in order to make gcc load the proper library, you need to explicitly tell it to in the call to compilation. gcc -lm testing.c -o math && ./testing

Notice the -lm flag? That tells gcc to load a specific library, and in this case, the math library.

;The procedure #[compiled-procedure 13 ("list" #x3) #x14 #x103498714] has been called with 1 argument; it requires exactly 2 arguments.

My code looked something like this:

(define (get-last-item lst)
  (define (recurse tsl l)
    (if (null? l)
      (car tsl)
      (recurse (cons (car l)) (cdr l))
    )
  )
  (recurse () lst)
)

In Scheme, it’s not so easy to realize you’ve missed an argument when you’re coming from a language where anything inside parenthesis are arguments. But in this case, it’s cons.

Why? It needs two arguments, something to car and something to cdr. That recurse line should look something more like this:

      (recurse (cons (car l) tsl) (cdr l))

Now, you may ask, “how did you know it was cons?” Aside from looking for something where I messed up arguments (1 instead of 2), I looked at the internal names. When you see something like #x14 #x103498714 in an error message, and as long as it’s a scheme procedure itself and not one of those fancy special forms, it will always have that representation internally. I ran cons point-blank in the Scheme interpreter and it returned the representation that shown in the error:

#[compiled-procedure 13 ("list" #x3) #x14 #x103498714]

I’m getting better at debugging Scheme code, but it’s still an uphill climb.

Happy cons’ing

;The procedure #[compiled-procedure 13 ("list" #x26) #x1a #x10349b102] has been called with 1 argument; it requires exactly 2 arguments.

One thing to notice: called with 1 argument; it requires exactly 2 arguments. That is a give away that you’ve missed argument somewhere, even if the code isn’t what is mentioned previously, as in this case, list.

  (cond
    ((< (distance (list-ref (- (length pt-list) 1) ) origin) delta-distance)
      (append pt-list p-list)
    )
    ((> (distance (list-ref 0) origin) delta-distance)
      (append p-list pt-list)
    )
    (else
      (append (car parts) p-list (cdr parts))  
    )
  )

Looking through that code, it isn’t really obvious what’s wrong. Distance takes two “points” and calculates the distance between them. Origin is point (0,0) and delta distance is previously calculated. The pt-list was a list of points to sort through and the p-list was a list-ized singular point for insertion. Confused yet? I was too.

Look at the innocent list-ref though? How many arguments does that have? The documentation says it has two arguments. I realized I missed the list argument in the call to list-ref.

Happy list-ref’ing.

Parse error: syntax error, unexpected T_AS, expecting ‘;’ in /homepages/224/d19925234/htdocs/kdelli/working/index.php on line 18

The error is so descriptive! Well, more so than others, actually. When do you use the keyword as? In loop declarations! See my code.

if ( $query == "places" ) {
    $output = "";
    for ($locations as $key => $l) {
        $output .= $l . "\r\n";
    }    
}

What could be wrong? Take a look at the for loop declaration. Notice anything odd about it? I didn’t either, at first. The as looks right, I took out $key => but there was no difference. I realized, since I was coming from Java, that for each loops aren’t just for loops in PHP. They are literally foreach loops.

I corrected the error simply by changing for to foreach.

;Ill-formed special form: (define sum-range a b count)

(define sum-range a b count)
 ; no procedure code yet
)

The error message mentions special form. If you’re learning Scheme, you should remember that define is a special form. It’s special in that it does not use the applicative ordering method for evaluation, among other things I suspect. So, somehow, the special form for define is incorrect. Any ideas? Can you spot the syntax error?

Notice that sum-range does not have a leading parenthesis? The procedure needs to be lead by a parenthesis so that it is its own expression (in terms of the special-formy-ness that is define). In short, the code should have read like so:

(define (sum-range a b count)
 ; no procedure code yet
)

It was really as easy as that.

The given code from the book is:

(define (factorial n)
  (if (= n 1)
      1
      (* n (factorial (- n 1)))))

I left this code as is. Factorials are easy to understand, pick any n-value and multiply each value below it until you get to 1.

Imagine you made a call to (factorial 2). The result will be 2 because of 1 * 2. Then imagine (factorial 1), and the result being 1 because the truthy-branch of the if-statement is used instead of the calculation branch. So far, so good, right?

Now, imagine you had a call to (factorial 0). What happens? You’re going to get a dreaded infinite recursion error, which will look similar to this:

;Aborting!: maximum recursion depth exceeded

What could cause this behavior? Perhaps the author of this method simply really trusted himself to not miscount in his recursion. Perhaps it was an oversight, but whatever the case, look at the condition that controls the procedure. Notice if (= n 1), and that there is only a single value that will make this procedure stop, and that is 1.

To fix this to accept 0, while not affecting the result of the factorial numerically, you can simply exchange the (= n 1) with (< n 1). This will force a 1 whenever n is less than 1, so that covers 0 and all negative values.

Happy factorial!